D, E and F are mid-points of the sides BC, CA and AB respectively of a ∆ABC. Prove that (i) FDCE is a parallelogram (ii) area of ∆DEF = area of ∆ABC (iii) area of || gm FDCE = area of ∆ABC

Question

KnowledgeBoat · Accepted Answer

∆ABC with D, E and F as mid-points of the sides BC, CA and AB, respectively is shown below: (i) F and E are midpoints of AB and AC respectively. So, by mid-point theorem, FE || BC and FE = BC .........(1) Also, D is the mid-point of BC CD = BC ........ (2) From (1) and (2), FE || BC and FE = CD Since, FE || BC so, FE || CD and FE = CD ......... (3) Similarly, D and F are the midpoints of BC and AB. So, by mid-point theorem, DF || AC and DF = AC .........(4) Also, E is the mid-point of AC EC = AC ........ (5) From (4) and (5), DF || AC and DF = EC Since, DF || AC so, DF || EC and DF = EC ......... (6) From 3 and 6, FE || CD, FE = CD, DF || EC and DF = EC. Since, opposite sides of FDCE are parallel and equal. Hence proved FDCE is a parallelogram. (ii) Since, BD = CD (As D is mid-point of BC), FE = CD (Proved above) and FE || CD So, BD = FE and BD || FE. Hence, BDEF is a || gm. Since, FD = EC (Proved above) and AE = EC (As E is mid-point of AC) and FD || EC. So, FD = AE and FD || AE. Hence, AFDE is a || gm. We know that, FDCE is a parallelogram and DE is a diagonal of || gm FDCE. So, area of ∆DEF = area of ∆DEC (As diagonal divides || gm into two triangles with equal area) ........(1) We know that, BDEF is a parallelogram and FD is a diagonal of || gm BDEF. So, area of ∆DEF = area of ∆FBD (As diagonal divides || gm into two triangles with equal area) ........(2) We know that, AFDE is a parallelogram and FE is a diagonal of || gm AFDE. So, area of ∆DEF = area of ∆AFE (As diagonal divides || gm into two triangles with equal area) ........(3) From 1, 2 and 3 we get, area of ∆DEF = area of ∆DEC = area of ∆FBD = area of ∆AFE .........(4) From figure, area of ∆ABC = area of ∆DEF + area of ∆DEC + area of ∆FBD + area of ∆AFE = area of ∆DEF + area of ∆DEF + area of ∆DEF + area of ∆DEF (From 4) = 4 x area of ∆DEF. ∴ area of ∆ABC = 4 x area of ∆DEF ⇒ area of ∆DEF = area of ∆ABC Hence, proved that area of ∆DEF = area of ∆ABC. (iii) From figure, Area of || gm FDCE = Area of ∆DEF + Area of ∆DEC = Area of ∆DEF + Area of ∆DEF (From part (ii) eqn. 4) = 2 area of ∆DEF = area of ∆ABC [As area of ∆DEF = area of ∆ABC] = area of ∆ABC. Hence, proved that area of || gm FDCE = area of ∆ABC.

Mathematics

D, E and F are mid-points of the sides BC, CA and AB respectively of a ∆ABC. Prove that (i) FDCE is a parallelogram (ii) area of ∆DEF = area of ∆ABC (iii) area of || gm FDCE = area of ∆ABC

Theorems on Area

Related Questions

In figure (1) given below, point D divides the side BC of ∆ABC in the ratio m : n. Prove that area of ∆ABD : area of ∆ADC = m : n.

In figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove that
area of ∆CPD = area of ∆AQD.

Mathematics

D, E and F are mid-points of the sides BC, CA and AB respectively of a ∆ABC. Prove that (i) FDCE is a parallelogram (ii) area of ∆DEF = area of ∆ABC (iii) area of || gm FDCE = area of ∆ABC

Theorems on Area

Related Questions

In figure (1) given below, point D divides the side BC of ∆ABC in the ratio m : n. Prove that area of ∆ABD : area of ∆ADC = m : n.

In figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove thatarea of ∆CPD = area of ∆AQD.

In figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove that
area of ∆CPD = area of ∆AQD.